printf上的分段错误-NASM 64bit Linux [英] Segmentation fault on printf - NASM 64bit Linux
问题描述
我试图使用scanf输入四个浮点数，将它们存储到堆栈中，然后使用vmovupd将它们复制到寄存器中以供使用.我的问题是，当我尝试输出这4个数字时，
I trying to input four floats using scanf, store them onto the stack, and then use vmovupd to copy them over to a register for use. My problem is when I try to output those 4 numbers, the program seg faults at printf.
我认为它与堆栈有关，但是我尝试多次弹出(一次执行多个指令)无济于事.我还是汇编语言的新手，所以使用gdb对我来说有点太高级了.
I presume it is something with the stack but I've tried popping numerous times (multiple instructions at once) to no avail. I'm still new to Assembly coding so using gdb is a bit too advanced for me.
您会注意到我包含了一个名为debug的文件.它使我可以查看寄存器和堆栈(这就是为什么有dumpstack指令的原因.)这是由我的教授提供的，它确实提供了一些帮助，但显然还不够(或者也许我只是缺少了一些东西).
You will notice that I have included a file called debug. It allows me to look at registers and the stack (that's why there's a dumpstack instruction.) That was provided by my professor and it did help some but obviously not enough (or maybe I am just missing something).
这是.cpp:
Here's the .cpp:
#include <iostream>
using namespace std;
extern "C" double ComputeElectricity();
int main()
{
cout << "Welcome to electric circuit processing by Chris Tarazi." << endl;
double returnValue = ComputeElectricity();
cout << "The driver received this number: " << returnValue << endl;
return 0;
}
这是ASM代码:
And here's the ASM code:
%include "debug.inc"
extern printf
extern scanf
global ComputeElectricity
;---------------------------------Declare variables-------------------------------------------
segment .data
greet db "This progam will help you analyze direct current circuits configured in parallel.", 10, 0
voltage db "Please enter the voltage of the entire circuit in volts: ", 0
first db "Enter the power consumption of device 1 (watts): ", 0
second db "Enter the power consumption of device 2 (watts): ", 0
third db "Enter the power consumption of device 3 (watts): ", 0
fourth db "Enter the power consumption of device 4 (watts): ", 0
thankyou db "Thank you. The computations have completed with the following results.", 10, 0
circuitV db "Curcuit total voltage: %1.18lf v", 10, 0
deviceNum db "Device number: 1 2 3 4", 10, 0
power db "Power (watts): %1.18lf %1.18lf %1.18lf %1.18lf", 10, 0
current db "Current (amps): %1.18lf %1.18lf %1.18lf %1.18lf", 10, 0
totalCurrent db "Total current in the circuit is %1.18lf amps.", 10, 0
totalPower db "Total power in the circuit is %1.18lf watts.", 10, 0
bye db "The analyzer program will now return total power to the driver.", 10, 0
string db "%s", 0
floatfmt db "%lf", 0
fourfloat db "%1.18lf %1.18lf %1.18lf %1.18lf", 0
;---------------------------------Begin segment of executable code------------------------------
segment .text
dumpstack 20, 10, 10
ComputeElectricity:
;dumpstack 30, 10, 10
;---------------------------------Output greet message------------------------------------------
mov qword rax, 0
mov rdi, string
mov rsi, greet
call printf
;---------------------------------Prompt for voltage--------------------------------------------
mov qword rax, 0
mov rdi, string
mov rsi, voltage
call printf
;---------------------------------Get voltage--------------------------------------------------
push qword 0
mov qword rax, 0
mov rdi, floatfmt
mov rsi, rsp
call scanf
vbroadcastsd ymm15, [rsp]
pop rax
;---------------------------------Prompt for watts 1--------------------------------------------
mov qword rax, 0
mov rdi, string
mov rsi, first
call printf
;---------------------------------Get watts 1---------------------------------------------------
push qword 0
mov qword rax, 0
mov rdi, floatfmt
mov rsi, rsp
call scanf
;---------------------------------Prompt for watts 2--------------------------------------------
mov qword rax, 0
mov rdi, string
mov rsi, second
call printf
;---------------------------------Get watts 2---------------------------------------------------
push qword 0
mov qword rax, 0
mov rdi, floatfmt
mov rsi, rsp
call scanf
;---------------------------------Prompt for watts 3--------------------------------------------
mov qword rax, 0
mov rdi, string
mov rsi, third
call printf
;---------------------------------Get watts 3---------------------------------------------------
push qword 0
mov qword rax, 0
mov rdi, floatfmt
mov rsi, rsp
call scanf
;---------------------------------Prompt for watts 4--------------------------------------------
mov qword rax, 0
mov rdi, string
mov rsi, fourth
call printf
;---------------------------------Get watts 4---------------------------------------------------
push qword 0
mov qword rax, 0
mov rdi, floatfmt
mov rsi, rsp
call scanf
;dumpstack 50, 10, 10
;---------------------------------Move data into correct registers------------------------------
vmovupd ymm14, [rsp] ; move all 4 numbers from the stack to ymm14
pop rax
pop rax
pop rax
pop rax
;dumpstack 55, 10, 10
vextractf128 xmm10, ymm14, 0 ; get lower half
vextractf128 xmm11, ymm14, 1 ; get upper half
;---------------------------------Move data into low xmm registers------------------------------
movsd xmm1, xmm11 ; move ymm[128-191] (3rd value) into xmm1
movhlps xmm0, xmm11 ; move from highest value from xmm11 to xmm0
movsd xmm3, xmm10
movhlps xmm2, xmm10
;showymmregisters 999
;---------------------------------Output results-------------------------------------------------
;dumpstack 60, 10, 10
mov rax, 4
mov rdi, fourfloat
push qword 0
call printf
pop rax
ret
推荐答案
问题出在堆栈使用上.
首先，ABI文档要求rsp在
First, the ABI docs mandate rsp be 16 byte aligned before a call.
由于call会将8字节的返回地址压入堆栈，因此您需要将rsp调整为16加8的倍数，以返回到16字节对齐方式. 16 * n + 8是包括的任何push指令或对RSP的其他更改，而不仅仅是sub rsp, 24.这是段错误的直接原因，因为printf将使用对齐的SSE指令，这将对未对齐的地址产生错误.
Since a call will push an 8 byte return address on the stack, you need to adjust rsp by a multiple of 16 plus 8 to get back to 16-byte alignment. The 16 * n + 8 is including any push instructions or other changes to RSP, not just sub rsp, 24. This is the immediate cause of the segfault, because printf will use aligned SSE instructions which will fault for unaligned addresses.
如果您解决了该问题，则堆栈仍然不平衡，因为您一直在推入值，但从不弹出它们.很难理解您要如何处理堆栈.
If you fix that, your stack is still unbalanced, because you keep pushing values but never pop them. It's hard to understand what you want to do with the stack.
通常的方法是在函数的开头(序言)中为本地人分配空间，并在函数的结尾(结语)中释放空间.如上所述，此数量(包括所有推送)应为16加8的倍数，因为函数 entry 上的RSP(在调用者的call之后)距离16字节边界8字节.
The usual way is to allocate space for the locals in the beginning of your function (the prologue) and free this at the end (epilogue). As discussed above, this amount (including any pushes) should be a multiple of 16 plus 8 because RSP on function entry (after the caller's call) is 8 bytes away from a 16-byte boundary.
在大多数glibc版本中，printf仅在AL！= 0时才关心16字节的堆栈对齐.(因为这意味着有FP args，因此它将所有XMM寄存器转储到堆栈中以便可以进行索引它们用于%f转换.)
In most builds of glibc, printf will only care about 16-byte stack alignment when AL != 0. (Because that means there are FP args, so it dumps all the XMM registers to the stack so it can index them for %f conversions.)
如果您使用未对齐的堆栈调用它仍然是一个错误，即使它恰好在您的系统上起作用；未来的glibc版本可以包含即使不使用FP args仍依赖16字节堆栈对齐的代码.例如，即使在大多数GNU/Linux发行版中，即使AL=0，scanf也确实会在未对齐的堆栈上崩溃.
It's still a bug if you call it with a misaligned stack even if it happens to work on your system; a future glibc version could include code that depends on 16-byte stack alignment even without FP args. For example, scanf already does crash on misaligned stacks even with AL=0 on most GNU/Linux distros.
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